Question

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.396235
R Square	0.157002
Adjusted R Square	0.156262
Standard Error	18.42647
Observations	1142
ANOVA
	df	SS	MS	F	Significance F
Regression	1	72088.71	72088.71	212.3161	3.12E-44
Residual	1140	387069.6	339.5348
Total	1141	459158.4
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	26.35917	0.803163	32.8192	7.4E-167	24.78333	27.93501	24.78333	27.93501
X Variable 1	0.000213	1.46E-05	14.57107	3.12E-44	0.000184	0.000242	0.000184	0.000242

a. Write the reqression equation.

2. Discuss the statistical significance of the model using the appropriate regression statistic at a 95% level of confidence.
3. Discuss the statistical significance of the coefficient for the independent variable using the appropriate regression statistic at a 95% level of confidence.
4. Interpret the coefficient for the independent variable.
5. What percentage of the observed variation in income is explained by the model?
6. Predict the value of a person’s income who works 50 hours a week, using this regression model.

Answer 1

Solution:

Given:

We are given regression analysis output of variables Income and Number of work hours per week.

Part a) Write the regression equation.

General regression equation is given by:

y = b₀ + b₁ * x

Here y = dependent variable = Income , x = independent variable = Number of work hours per week

b0 = Intercept = 26.35917

b1 = Slope = 0.000213

Thus regression equation is:

Income = 26.35917 + 0.000213 * Number of work hours per week

Part b) Discuss the statistical significance of the model using the appropriate regression statistic at a 95% level of confidence , that is for 1 - 0.95 = 0.05 level of significance.

ANOVA table is used to test the significance of overall model.

From ANOVA we can see: Significance F = 3.12E-44

3.12E-44 is scientific number and E-44 means we have to move 44 decimal places to left of 3.12

Thus rounding this number to 4 decimal places we get:

Significance F = 0.0000

That is: P-value = 0.0000

Since P-value = 0.0000 < 0.05 level of significance , we conclude that the fitted model is significant.

Part c) Discuss the statistical significance of the coefficient for the independent variable using the appropriate regression statistic at a 95% level of confidence.

From P-value column, we can see P-value for slope coefficient is 3.12E-44 = 0.0000

Since P-value = 0.0000 < 0.05 level of significance , the coefficient for the independent variable is statistically significant.

Part d) Interpret the coefficient for the independent variable.

Intercept is the value of dependent variable y , when x = 0.

Here y is income and x is Number of work hours per week.

If Number of work hours per week is 0, then there would be no possibility of Income.

Thus intercept is meaningless in this context.

That is: if Number of work hours per week is 0 , then it is meaningless to have income of 26.35917 .

Slope: Slope is the coefficient of independent variable. It gives amount of change in dependent variable when independent variable changed by one unit.

Thus we have Slope = 0.000213.

Thus if we increase number of hour per week by one hour, then there is increase of Income by the amount 0.000213.

Part e) What percentage of the observed variation in income is explained by the model?

R-square gives the amount variation in dependent variable which is explained by model ( by independent variable)

Here R-square = 0.157002 = 15.7%

Thus About 15.7% of the observed variation in income is explained by the model.

Part f) Predict the value of a person’s income who works 50 hours a week, using this regression model.

Put Number of work hours per week = 50 in regression equation obtained in part a)

Income = 26.35917 + 0.000213 * Number of work hours per week

Income = 26.35917 + 0.000213 * 50

Income = 26.35917 + 0.01065

Income = 26.36982

Thus the predicted value of a person’s income who works 50 hours a week is 26.36982